Problem Set
33: | 1
| 2 |
3 | 4 | 5 | 6
| 9 | 24 | 25 | 26
| 28 | 29 | 32 | 36
| 37 | Go up

** - by Zach Peterson, 2003**

**1.
The parallax angle of a star is 0.00017 degrees. How far away is the star?**

d (parsec) = 1/p (arc-second)

0.00017 x 3600 = .612 arc-seconds

1/.612 = 1.634 parsecs

1.634 x 3.26 =

5.3 ly

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**2.
A star exhibits a parallax of 0.28 arc-seconds. How far away is it?**

d (parsec) = 1/p (arc-second)

d = 1/.28 = 3.57 parsecs

3.57 x 3.26 =

11.64 ly

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**3.
Using the definitions of the parsec and the light-year, show that 1 pc = 3.26 ly. **

1 parsec = 3.26 ly

1 parsec = 1 A.U./tan(2.8 x 10^-4) = 3.086 x 10^13 km = (3.086 x 10^13)(9.46 x 10^12 km/ly) =

3.26 ly

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**4.
A star is 24 parsecs away. What is its parallax angle? State (a) in seconds of
arc, and (b) in degrees.**

(a) 1/24 =

.0416 arc seconds(b) .0416 x (1/3600) =

1.16 x 10^-5 degrees

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**5.
What is the parallax angle of a star if it is 42 light-years away? How many
parsecs is this? **

42 ly/3.26 =

13 parsecs1/13 =

.077 arc seconds

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**6.
A star is 25 parsecs away. How long does it take for its light to reach us?**

Convert parsecs to light-years.

25 x 3.26 = 81.5 ly =

81.5 years

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**9.
What is the apparent brightness of the sun as seen on Jupiter? (Jupiter is 5.2
AU away)**

b = L/(4pd

^{2})d = 7.783 x 10^6

L = 8.67 x 10^-8(4pr

^{2})T^{4}L = 3.68 x 10^20

b = 3.68 x 10^20/(4p7.8 x 10^8

^{2})b = 48.1 W/m

^{2}

**24.
If a galaxy is traveling away from us at 1.0 percent of the speed of light,
roughly how far away is it?**

v = Hd

3000

= 80dkm/sd = 3.75 x 3.26 x 10^6 =

1.2 x 10^8 ly

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**25.
The redshift of a galaxy indicates a velocity of 2500 km/s. How far away is it?**

v = Hd <-

Mpc!!!

2500 = 80dd =

31.25 Mpc

**26.
Estimate the speed of a galaxy (relative to us) that is near the
"edge" of the universe, say 10-billion light-years away.**

Convert light-years to Mpc.

10^9/3.26 x 10^6 = 1,000,000 Mpc

v = 80 x 1,000,000

v =

.8c

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**28.
Estimate the observed wavelength for the 656-nm line in the Balmer series of
hydrogen in the spectrum of a galaxy whose distance from us is (a) 10 ^{6}
ly, (b) 10^{8} ly, (c) 10^{10} ly**

The first thing to do is turn the LY into Mpc. Since 1 pc is 3.26 LY, then

10^{6}Ly = .306748 Mpc

10^{8}Ly = 30.6748 Mpc and finally

10^{10}Ly = 3067.48 Mpc

If you use v = Hd to get the recession velocity, then you arrive at these recession veolocities

10^{6}Ly = .306748 Mpc --> 24.54 km/s

10^{8}Ly = 30.6748 Mpc --> 2454 km/s

10^{10}Ly = 3067.48 Mpc --> 245,400 km/s

Since the speed of light is 300,000 km/s the approximate formula yields

(a) Dl/l = v/c

Dl/656 nm = 24.54 km/s/300,000 km/s, Dl = .054 nm, = 656.054

(b) Dl/l = v/c

Dl/656 nm = 2454 km/s/300,000 km/s, Dl = 5.4 nm, = 661.4

(c) Well - 245,400 km/s is not small compared to 300,000 km/s (i.e. v is not << c)

So you will have to use equation 33-3 from the book (The real equation):I get 2073 nm when I do it, they rounded a bit and got 2170 nm I think I am right

- but since the Hubble constant is largely uncertain...

**29.
Estimate the speed of a galaxy, and its distance from us, if the wavelength for
the hydrogen line at 434 nm is measured on Earth as being 610 nm?**

use the formulae m = m

_{0}/g and g = 1/Ö1 - (v^{2}/c^{2})

using the rules of perspective and simultaneity, we know that l' = 610 nm and that l = 434 nm610 = 434Ö1 - (v

^{2}/c^{2})Do the math and you get

0.328cNow to get the distance you use v = Hd

.328c = 80d

d =

4 x 10^9 ly

**32.
Calculate the wavelength at the peak of the blackbody radiation distribution at
2.7 K using the Wien law.**

l(2.7) = 2.9 x 10^-3

l = .001074

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**36.
Suppose that three main-sequence stars could undergo the three changes
represented by the three arrows, A, B, and C, in the H-R diagram of Fig. 33-26.
For each case, describe the changes in temperature, luminosity, and size.**

<Fig 33-26 on page 1034>

A:

T increases, L constant, size decreasesB:

T constant, L decreases, size decreasesC:

T decreases, L increases, size increases

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**37.
Assume that the nearest stars to us have an intrinsic luminosity about the same
as the Sun's. Their apparent brightness, however, is about 10^11 times fainter
than the Sun. From this, estimate the distance to the nearest stars. (Newton did
this calculation, although he made a numerical error of a factor of 100)**

b(sun) = 10^11 b(stars)

L(sun)/d^2 = 10^11 L(stars)/4pd^2

(1.5 x 10^11)^2 = 10^11/(1.5 x 10^11)^2

5.0 ly

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